Mathematics Higher level - aptutorgroup.com · Math higher level paper 3 statistics and probability specimen question paper Math higher level paper 3 statistics and probability specimen

Mathematics Higher level Specimen papers 1, 2 and 3 (adapted from November 2014) For first examinations in 2017

CONTENTS Math higher level paper 1 specimen question paper Math higher level paper 1 specimen markscheme Math higher level paper 2 specimen question paper Math higher level paper 2 specimen markscheme Math higher level paper 3 discrete specimen question paper Math higher level paper 3 discrete specimen markscheme Math higher level paper 3 calculus specimen question paper Math higher level paper 3 calculus specimen markscheme Math higher level paper 3 sets, relations and groups specimen question paper Math higher level paper 3 sets, relations and groups specimen markscheme Math higher level paper 3 statistics and probability specimen question paper Math higher level paper 3 statistics and probability specimen markscheme

ApTutorGroup.com

ApTutorGroup.com

Candidate session number

SPEC/5/MATHL/HP1/ENG/TZ0/XX

MathematicsHigher levelPaper 1

© International Baccalaureate Organization 201612 pages

Instructions to candidates

yy Write your session number in the boxes above.yy Do not open this examination paper until instructed to do so.yy You are not permitted access to any calculator for this paper.yy Section A: answer all questions. Answers must be written within the answer boxes provided.yy Section B: answer all questions in the answer booklet provided. Fill in your session number

on the front of the answer booklet, and attach it to this examination paper and your cover sheet using the tag provided.

yy Unless otherwise stated in the question, all numerical answers should be given exactly or correct to three significant figures.yy A clean copy of the mathematics HL and further mathematics HL formula booklet is

required for this paper.yy The maximum mark for this examination paper is [100 marks].

2 hours

SPECIMEN (adapted from November 2014)

12EP01

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ApTutorGroup.com

– 2 –

Full marks are not necessarily awarded for a correct answer with no working. Answers must be supported by working and/or explanations. Where an answer is incorrect, some marks may be given for a correct method, provided this is shown by written working. You are therefore advised to show all working.

Section A

Answer all questions. Answers must be written within the answer boxes provided. Working may be continued below the lines, if necessary.

1. [Maximum mark: 4]

The function f is defined by f xx

( ) =1 , x ≠ 0 .

The graph of the function y = g (x) is obtained by applying the following transformations to the graph of y = f (x) :

a translation by the vector −

30

;

a translation by the vector 01

.

(a) Find an expression for g (x) . [2]

(b) State the equations of the asymptotes of the graph of g . [2]

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12EP02

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– 3 –

Turn over


The quadratic equation 2x2 - 8x + 1 = 0 has roots α and β .

(a) Without solving the equation, find the value of

(i) α + β ;

(ii) αβ . [2]

Another quadratic equation x2 + px + q = 0 , p , q ∈ has roots 2α

and 2β

.

(b) Find the value of p and the value of q . [4]

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12EP03


ApTutorGroup.com

– 4 –


A point P, relative to an origin O, has position vector OP→

=++−

13 21

ss

s, s ∈ .

(a) Show that OP→

= + +2

26 12 11s s . [1]

(b) Hence find the minimum length of OP→

. [4]

(c) Explain, geometrically, why your answer gives a minimum value. [1]

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12EP04


ApTutorGroup.com

– 5 –

Turn over


Events A and B are such that P(A) = 0.2 and P(B) = 0.5 .

(a) Determine the value of P(A ∪ B) when

(i) A and B are mutually exclusive;

(ii) A and B are independent. [4]

(b) Find the smallest and largest possible values of P(A | B) . [3]

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12EP05


ApTutorGroup.com

– 6 –


By using the substitution u x= +1 , find xx

x1+

d∫ .

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12EP06


ApTutorGroup.com

– 7 –

Turn over


Use mathematical induction to prove that (2n)! ≥ 2n(n!)2 , n ∈ + .

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12EP07


ApTutorGroup.com

– 8 –


A continuous random variable T has probability density function f defined by

f tt t

( ),,

=−

20 otherwise

1≤ ≤ 3.

(a) Sketch the graph of y = f (t) . [2]

(b) (i) Find the lower quartile of T .

(ii) Hence find the interquartile range of T . [5]

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12EP08


ApTutorGroup.com

– 9 –

Turn over


A set of positive integers {1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9} is used to form a pack of nine cards. Each card displays one positive integer without repetition from this set. Grace wishes to select four cards at random from this pack of nine cards.

(a) Find the number of selections Grace could make if the largest integer drawn among the four cards is either a 5, a 6 or a 7. [3]

(b) Find the number of selections Grace could make if at least two of the four integers drawn are even. [4]

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12EP09


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– 10 –

Do not write solutions on this page.

Section B

Answer all questions in the answer booklet provided. Please start each question on a new page.


The function f is defined as f (x) = e3x+1 , x ∈ .

(a) Find f -1(x) . [3]

The function g is defined as g(x) = ln x , x ∈ + . The graph of y = g (x) intersects the x-axis at the point Q.

(b) Show that the equation of the tangent T to the graph of y = g(x) at the point Q is y = x - 1 . [3]

A region R is bounded by the graphs of y = g(x) , the tangent T and the line x = e .

(c) Find the area of the region R . [5]

(d) (i) Show that g(x) ≤ x - 1 , x ∈ + .

(ii) By replacing x with 1x

in part (d)(i), show that x

xg x−1 ( )≤ , x ∈ + . [6]

12EP10


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– 11 –

Turn over



The position vectors of the points A, B and C are a , b and c respectively, relative to an origin O. The following diagram shows the triangle ABC and points M, R, S and T.

A

R

MTS

C B

diagram not to scale

M is the mid-point of [AC].

R is a point on [AB] such that AR AB→ →

=13

.

S is a point on [AC] such that AS AC→ →

=23

.

T is a point on [RS] such that RT RS→ →

=23

.

(a) (i) Express AM→

in terms of a and c .

(ii) Hence show that BM→

= − +12

12

a b c . [4]

(b) (i) Express RA→

in terms of a and b .

(ii) Show that RT→

= − − +29

29

49

a b c . [5]

(c) Prove that T lies on [BM]. [5]

12EP11


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– 12 –



(a) Show that ( tan ) ( tan ) coscos

, cos1 1 2+ + − =i iθ θ

θθ

θn nn

n ≠ 0 . [6]

(b) (i) Use the double angle identity tan tantan

2 21 2θ

θθ

=−

to show that tan 8

2 1= −π .

(ii) Show that cos 4x = 8 cos4 x - 8 cos2 x + 1 .

(iii) Hence find the value of 2 420

8 coscos

xx

xd∫π

. [13]

12EP12


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SPEC/5/MATHL/HP1/ENG/TZ0/XX/M

16 pages

Markscheme

Specimen (adapted from November 2014)

Mathematics

Higher level

Paper 1

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– 2 – SPEC/5/MATHL/HP1/ENG/TZ0/XX/M

Instructions to Examiners

Abbreviations M Marks awarded for attempting to use a valid Method; working must be seen. (M) Marks awarded for Method; may be implied by correct subsequent working.

A Marks awarded for an Answer or for Accuracy; often dependent on preceding M marks.

(A) Marks awarded for an Answer or for Accuracy; may be implied by correct subsequent working. R Marks awarded for clear Reasoning. N Marks awarded for correct answers if no working shown. AG Answer given in the question and so no marks are awarded.

Using the markscheme 1 General

Mark according to RM™ Assessor instructions and the document “Mathematics HL: Guidance for e-marking May 2016”. It is essential that you read this document before you start marking. In particular, please note the following. Marks must be recorded using the annotation stamps. Please check that you are entering marks for the right question. If a part is completely correct, (and gains all the ‘must be seen’ marks), use the ticks with

numbers to stamp full marks. If a part is completely wrong, stamp A0 by the final answer. If a part gains anything else, it must be recorded using all the annotations. All the marks will be added and recorded by RM™ Assessor.

2 Method and Answer/Accuracy marks

Do not automatically award full marks for a correct answer; all working must be checked, and marks awarded according to the markscheme.

It is not possible to award M0 followed by A1, as A mark(s) depend on the preceding M mark(s), if any.

Where M and A marks are noted on the same line, for example, M1A1, this usually means M1 for an attempt to use an appropriate method (for example, substitution into a formula) and A1 for using the correct values.

Where the markscheme specifies (M2), N3, etc, do not split the marks.

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Once a correct answer to a question or part-question is seen, ignore further correct working. However, if further working indicates a lack of mathematical understanding do not award the final A1. An exception to this may be in numerical answers, where a correct exact value is followed by an incorrect decimal. However, if the incorrect decimal is carried through to a subsequent part, and correct FT working shown, award FT marks as appropriate but do not award the final A1 in that part.

Examples

Correct answer seen Further working seen Action 1.

8 2 5.65685… (incorrect decimal value)

Award the final A1 (ignore the further working)

2. 1sin 4

4x sin x Do not award the final A1

3. log – loga b log ( – )a b Do not award the final A1

3 N marks Award N marks for correct answers where there is no working.

Do not award a mixture of N and other marks. There may be fewer N marks available than the total of M, A and R marks; this is deliberate as it

penalizes candidates for not following the instruction to show their working.

4 Implied marks Implied marks appear in brackets, for example, (M1), and can only be awarded if correct work is

seen or if implied in subsequent working. Normally the correct work is seen or implied in the next line. Marks without brackets can only be awarded for work that is seen.

5 Follow through marks Follow through (FT) marks are awarded where an incorrect answer from one part of a question is

used correctly in subsequent part(s). To award FT marks, there must be working present and not just a final answer based on an incorrect answer to a previous part. If the question becomes much simpler because of an error then use discretion to award fewer

FT marks. If the error leads to an inappropriate value (for example, sin 1.5 ), do not award the mark(s)

for the final answer(s). Within a question part, once an error is made, no further dependent A marks can be awarded,

but M marks may be awarded if appropriate. Exceptions to this rule will be explicitly noted on the markscheme.

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6 Misread

If a candidate incorrectly copies information from the question, this is a misread (MR). A candidate should be penalized only once for a particular misread. Use the MR stamp to indicate that this has been a misread. Then deduct the first of the marks to be awarded, even if this is an M mark, but award all others so that the candidate only loses one mark. If the question becomes much simpler because of the MR, then use discretion to award fewer

marks. If the MR leads to an inappropriate value (for example, sin 1.5 ), do not award the mark(s) for

the final answer(s).

7 Discretionary marks (d) An examiner uses discretion to award a mark on the rare occasions when the markscheme does

not cover the work seen. In such cases the annotation DM should be used and a brief note written next to the mark explaining this decision.

8 Alternative methods

Candidates will sometimes use methods other than those in the markscheme. Unless the question specifies a method, other correct methods should be marked in line with the markscheme. If in doubt, contact your team leader for advice.

Alternative methods for complete questions are indicated by METHOD 1, METHOD 2, etc. Alternative solutions for part-questions are indicated by EITHER… OR. Where possible, alignment will also be used to assist examiners in identifying where these

alternatives start and finish.

9 Alternative forms

Unless the question specifies otherwise, accept equivalent forms.

As this is an international examination, accept all alternative forms of notation. In the markscheme, equivalent numerical and algebraic forms will generally be written in

brackets immediately following the answer. In the markscheme, simplified answers, (which candidates often do not write in examinations),

will generally appear in brackets. Marks should be awarded for either the form preceding the bracket or the form in brackets (if it is seen).

Example: for differentiating ( ) 2sin (5 3)f x x , the markscheme gives: ( ) 2cos(5 3) 5f x x 10cos(5 3)x A1

Award A1 for 2cos(5 3) 5x , even if 10cos (5 3)x is not seen.

10 Accuracy of Answers

Candidates should NO LONGER be penalized for an accuracy error (AP). If the level of accuracy is specified in the question, a mark will be allocated for giving the answer to

the required accuracy. When this is not specified in the question, all numerical answers should be given exactly or correct to three significant figures. Please check work carefully for FT.

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11 Crossed out work

If a candidate has drawn a line through work on their examination script, or in some other way crossed out their work, do not award any marks for that work.

12 Calculators

No calculator is allowed. The use of any calculator on paper 1 is malpractice, and will result in no grade awarded. If you see work that suggests a candidate has used any calculator, please follow the procedures for malpractice. Examples: finding an angle, given a trig ratio of 0.4235.

13 More than one solution

Where a candidate offers two or more different answers to the same question, an examiner should only mark the first response unless the candidate indicates otherwise.

14 Candidate work

Candidates are meant to write their answers to Section A on the question paper (QP), and Section B on answer booklets. Sometimes, they need more room for Section A, and use the booklet (and often comment to this effect on the QP), or write outside the box. This work should be marked. The instructions tell candidates not to write on Section B of the QP. Thus they may well have done some rough work here which they assume will be ignored. If they have solutions on the answer booklets, there is no need to look at the QP. However, if there are whole questions or whole part solutions missing on answer booklets, please check to make sure that they are not on the QP, and if they are, mark those whole questions or whole part solutions that have not been written on answer booklets.

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Section A 1. (a) 1

( ) 13

g xx

A1A1

Note: Award A1 for 3x in the denominator and A1 for the “ 1 ”. [2 marks] (b) 3x A1 1y A1 [2 marks] Total [4 marks] 2. (a) using the formulae for the sum and product of roots: (i) 4 A1

(ii) 1

2 A1

Note: Award A0A0 if the above results are obtained by solving

the original equation (except for the purpose of checking). [2 marks] (b) METHOD 1

required quadratic is of the form 2 2 2 2 2x x

(M1)

4q

8q A1

2 2p

2( )

M1

2 412

16p A1

Note: Accept the use of exact roots.

continued…

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Question 2 continued METHOD 2

replacing x with 2

x M1

2

2 22 8 1 0

x x

2

8 161 0

xx (A1)

2

16 8 0x x 16p and 8q A1A1

Note: Award A1A0 for 2 16 8 0x x ie, if 16p and 8q are not explicitly stated.

[4 marks] Total [6 marks]

3. (a) 2

2 2 2OP (1 ) (3 2 ) (1 )s s s

A1

26 12 11s s AG [1 mark]

(b) attempt to differentiate: 2

OP ( 12 12)d sds

M1

attempt to solve: 2

OP 0 for d sds

(M1)

1s A1

the minimum length of OP

is 5 A1 [4 marks] (c) (The point P is restricted to a line) a line has a unique point of closest approach to the origin, but no maximum R1 [1 mark] Total [6 marks]

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4. (a) (i) use of P ( ) P ( ) P ( )A B A B (M1) P ( ) 0.2 0.5A B 0.7 A1 (ii) use of P ( ) P ( ) P ( ) P ( )P ( )A B A B A B (M1) P ( ) 0.2 0.5 0.1A B 0.6 A1 [4 marks]

(b) P( )

P( | )P( )

A BA BB

P( | )A B is a minimum when A B P( ) 0A B R1

P( | )A B is a maximum when A B P( ) P( )A B A R1

min value 0, max value 0.4 A1 [3 marks] Total [7 marks]

5. d 1

d 2

ux x A1

d 2 ( 1) dx u u Note: Award the A1 for any correct relationship between dx and du.

2( 1)

d 2 d1

x ux uux

(M1)A1

Note: Award the M1 for an attempt at substitution resulting in an integral only involving u.

12 2 du u

u (A1)

2 4 2ln ( )u u u C A1

2 3 2 ln 1 ( )x x x C A1 Note: Award the A1 for a correct expression in x, but not necessarily fully expanded/simplified.

[6 marks]

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6. let P ( )n be the proposition that 2(2 )! 2 ( !)nn n , n

consider P (1) :

2! 2 and 212 1! 2 so P (1) is true R1

assume P ( )k is true ie 2(2 )! 2 ( !)kk k , k

M1

Note: Do not award M1 for statements such as “let n k ”.

consider P ( 1)k :

2( 1) ! (2 2)(2 1)(2 )!k k k k M1

22( 1) ! (2 2)(2 1)( !) 2kk k k k A1 Note: Condone “working backwards” up to this point, but no further unless it is fully justified.

22( 1)(2 1)( !) 2kk k k

1 22 ( 1)( 1)( !)k k k k since 2 1 1k k R1

212 ( 1)!k k A1

P ( 1)k is true whenever P ( )k is true and P (1) is true, so P ( )n is true for n R1 Note: To obtain the final R1, four of the previous marks must have been awarded.

[7 marks] 7. (a)

2 t correct for [1, 2] A1

2 t correct for [2 , 3] A1 [2 marks]

(b) (i) let 1q be the lower quartile

consider 1

1

1(2 )

4

q

t dt M1A1

obtain 1

12

2q A1

continued…

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Question 7 continued

(ii) by symmetry, for example, 3

12

2q A1

hence IQR 2 A1

Note: Only accept this final answer for the A1. [5 marks] Total [7 marks] 8. (a) use of the addition principle with 3 terms (M1)

to obtain 4 5 63 3 3C C C ( 4 10 20) A1

number of possible selections is 34 A1 [3 marks] (b) EITHER

recognition of three cases: (2 odd and 2 even or 1 odd and 3 even or 0 odd and 4 even) (M1)

5 4 5 4 5 42 2 1 3 0 4C C C C C C ( 60 20 1) (M1)A1

OR

recognition to subtract the sum of 4 odd and 3 odd and 1 even from the total (M1)

9 5 5 44 4 3 1C C C C ( 126 5 40) (M1)A1

THEN

number of possible selections is 81 A1 [4 marks] Total [7 marks]

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Section B 9. (a) 3 1e yx M1

Notes: The M1 is for switching variables and can be awarded at any stage. Further marks do not rely on this mark being awarded.

taking the natural logarithm of both sides and attempting to transpose M1

1 1( ) (ln 1)

3f x x A1

[3 marks] (b) coordinates of Q are (1, 0) seen anywhere A1

d 1

d

yx x M1

at Q, d1

d

yx A1

1y x AG [3 marks] (c) let the required area be A

1 1

1d ln de e

A x x x x M1

Notes: The M1 is for a difference of integrals. Condone absence of limits here.

attempting to use integration by parts to find ln dx x (M1)

e2e1

1

[ ln ]2

x x x x x

A1A1

Note: Award A1 for 2

–2

x x and A1 for ln –x x x .

Note: The second M1 and second A1 are independent of the first M1 and the first A1.

2 2

e 1 e 2e 1e

2 2 2

A1

[5 marks]

continued…

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Question 9 continued (d) (i) METHOD 1 consider for example ( ) 1 lnh x x x

(1) 0h and 1( ) 1h x

x (A1)

as ( ) 0h x for 1x , then ( ) 0h x for 1x R1

as ( ) 0h x for 0 1x , then ( ) 0h x for 0 1x R1

so ( ) 1g x x , x AG METHOD 2

2

1( )g x

x A1

( ) 0g x (concave down) for x R1

the graph of ( )y g x is below its tangent ( 1y x at 1x ) R1

so ( ) 1g x x , x AG

Note: The reasoning may be supported by drawn graphical arguments. METHOD 3

clear correct graphs of 1y x and ln x for 0x A1A1

statement to the effect that the graph of ln x is below the graph of its tangent at 1x R1AG

continued…

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(ii) replacing x by 1

x to obtain

1 1 1ln 1

xx x x

M1

1 1

ln 1xx

x x

(A1)

1 1

ln 1xx

x x

A1

so 1( )

x g xx

, x AG


10. (a) (i) 1AM AC

2

(M1)

1( )

2 c a A1

(ii) BM BA AM

M1

1( )

2 a b c a A1

1 1BM

2 2

a b c AG

[4 marks]

(b) (i) 1RA BA

3

1( )

3 a b A1

(ii) 2RT RS

3

2

RA AS3

(M1)

2 1 2

( ) ( )3 3 3

a b c a or equivalent A1A1

2 4( ) ( )

9 9 a b c a A1

2 2 4RT

9 9 9

a b c AG

[5 marks]

continued…

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(c) BT BR RT

2BA RT

3

(M1)

2 2 2 2 4

3 3 9 9 9 a b a b c A1

8 1 1

BT9 2 2

a b c A1

point B is common to BT

and BM

and 8BT BM

9

R1R1

so T lies on [BM] AG [5 marks] Total [14 marks] 11. (a) METHOD 1

sin sin(1 i tan ) (1 i tan ) 1 i 1 i

cos cos

n nn n

M1

cos i sin cos i sin

cos cos

n n

A1

by de Moivre’s theorem (M1)

cos i sin

cos

n

cos i sin

cosn

n n

A1

recognition that cos isin is the complex conjugate of cos isin (R1)

use of the fact that the operation of complex conjugation commutes with the operation of raising to an integer power

cos i sin

cos

n

cos i sin

cos n

n n

A1

2 cos(1 i tan ) (1 i tan )

cosn n

n

n

AG

continued…

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Question 11 continued METHOD 2

(1 itan ) (1 itan ) (1 itan ) (1 itan(– ))n n n n (M1)

cos ( ) i sin ( )(cos i sin )

cos cos

nn

n n

M1A1

Note: Award M1 for converting to cosine and sine terms.

use of de Moivre’s theorem (M1)

1cos i sin cos ( ) i sin ( )

cosn n n n n

A1

2 cos

cos n

n

as cos ( ) cosn n and sin ( ) sinn n R1AG

[6 marks]

(b) (i) 2

π2 tanπ 8tan

π4 1 tan8

(M1)

2 π πtan 2 tan 1 0

8 8 A1

let πtan

8t

attempting to solve 2 2 1 0t t for t M1 1 2t A1

π

8 is a first quadrant angle and tan is positive in this quadrant,

so πtan 0

8 R1

πtan 2 1

8 AG

(ii) 2cos 4 2 cos 2 – 1x x A1 222 2 cos – 1 – 1x M1

4 22 4cos – 4cos 1 –1x x A1

4 28 cos – 8 cos 1x x AG

Note: Accept equivalent complex number derivation.

continued…

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(iii) π π 4 28 8

2 20 0

2cos4 8cos 8cos 1d 2 d

cos cos

x x xx xx x

π

2 28

02 8 cos 8 sec dx x x M1

Note: The M1 is for an integrand involving no fractions.

use of 2 1cos (cos 2 1)

2x x M1

π

28

02 4 cos 2 4 sec dx x x A1

π

80

4sin2 8 2tanx x x A1

4 2 π 2 (or equivalent) A1 [13 marks] Total [19 marks]

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Candidate session number

SPEC/5/MATHL/HP2/ENG/TZ0/XX

MathematicsHigher levelPaper 2



yy Write your session number in the boxes above.yy Do not open this examination paper until instructed to do so.yy A graphic display calculator is required for this paper.yy Section A: answer all questions. Answers must be written within the answer boxes provided.yy Section B: answer all questions in the answer booklet provided. Fill in your session number

on the front of the answer booklet, and attach it to this examination paper and your cover sheet using the tag provided.

yy Unless otherwise stated in the question, all numerical answers should be given exactly or correct to three significant figures.yy A clean copy of the mathematics HL and further mathematics HL formula booklet is

required for this paper. yy The maximum mark for this examination paper is [100 marks].

2 hours


16EP01

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– 2 –

Full marks are not necessarily awarded for a correct answer with no working. Answers must be supported by working and/or explanations. In particular, solutions found from a graphic display calculator should be supported by suitable working. For example, if graphs are used to find a solution, you should sketch these as part of your answer. Where an answer is incorrect, some marks may be given for a correct method, provided this is shown by written working. You are therefore advised to show all working.

Section A

Answer all questions. Answers must be written within the answer boxes provided. Working may be continued below the lines, if necessary.


Consider the two planes

π1 : 4x + 2y - z = 8

π2 : x + 3y + 3z = 3.

Find the angle between π1 and π2 , giving your answer correct to the nearest degree.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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16EP02


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– 3 –

Turn over


The wingspans of a certain species of bird can be modelled by a normal distribution with mean 60.2 cm and standard deviation 2.4 cm.

According to this model, 99 % of wingspans are greater than x cm.

(a) Find the value of x . [2]

In a field experiment, a research team studies a large sample of these birds. The wingspans of each bird are measured correct to the nearest 0.1 cm.

(b) Find the probability that a randomly selected bird has a wingspan measured as 60.2 cm. [4]

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16EP03


ApTutorGroup.com

– 4 –


The lines l1 and l2 are defined as

l x y z1

13

52

122

: −=

−=

−−

l x y z2

18

511

126

: −=

−=

−.

The plane π contains both l1 and l2 .

(a) Find the Cartesian equation of π . [4]

The line l3 passing through the point (4 , 0 , 8) is perpendicular to π .

(b) Find the coordinates of the point where l3 meets π . [4]

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16EP04


ApTutorGroup.com

– 5 –

Turn over


Consider p(x) = 3x3 + ax + 5a , a ∈ .

The polynomial p(x) leaves a remainder of -7 when divided by (x - a) .

Show that only one value of a satisfies the above condition and state its value.

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16EP05


ApTutorGroup.com

– 6 –


The seventh, third and first terms of an arithmetic sequence form the first three terms of a geometric sequence.

The arithmetic sequence has first term a and non-zero common difference d .

(a) Show that d a=

2. [3]

The seventh term of the arithmetic sequence is 3. The sum of the first n terms in the arithmetic sequence exceeds the sum of the first n terms in the geometric sequence by at least 200.

(b) Find the least value of n for which this occurs. [6]

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16EP06


ApTutorGroup.com

– 7 –

Turn over


A particle moves in a straight line such that its velocity, v m s–1 , at time t seconds, is given by

v tt t

t t( )

( ) ,

,=

− −

− >

5 2 0 4

32

4

2 ≤≤.

(a) Find the value of t when the particle is instantaneously at rest. [2]

The particle returns to its initial position at t = T .

(b) Find the value of T . [5]

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16EP07


ApTutorGroup.com

– 8 –


Compactness is a measure of how compact an enclosed region is.

The compactness, C , of an enclosed region can be defined by CAd

=4

2π, where A is the area

of the region and d is the maximum distance between any two points in the region.

For a circular region, C = 1 .

Consider a regular polygon of n sides constructed such that its vertices lie on the circumference of a circle of diameter x units.

(a) If n > 2 and even, show that C nn

=2

2sin ππ

. [3]

If n > 1 and odd, it can be shown that Cn

n

n

=+

sin

cos

2

1 π

π

π.

(b) Find the regular polygon with the least number of sides for which the compactness is more than 0.99. [4]

(c) Comment briefly on whether C is a good measure of compactness. [1]

16EP08


ApTutorGroup.com

– 9 –

Turn over

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16EP09


ApTutorGroup.com

– 10 –


Section B

Answer all questions in the answer booklet provided. Please start each question on a new page.


Consider the triangle PQR where ˆQPR 30= �, PQ = (x + 2) cm and PR = (5 - x)2 cm , where -2 < x < 5 .

(a) Show that the area, A cm2 , of the triangle is given by A x x x= − + +( )14

8 5 503 2 . [2]

(b) (i) State ddAx

.

(ii) Verify that ddAx

= 0 when x =13

. [3]

(c) (i) Find dd

2

2

Ax

and hence justify that x =13

gives the maximum area of triangle PQR.

(ii) State the maximum area of triangle PQR.

(iii) Find QR when the area of triangle PQR is a maximum. [7]

16EP10


ApTutorGroup.com

– 11 –

Turn over



The number of complaints per day received by customer service at a department store follows a Poisson distribution with a mean of 0.6.

(a) On a randomly chosen day, find the probability that

(i) there are no complaints;

(ii) there are at least three complaints. [3]

(b) In a randomly chosen five-day week, find the probability that there are no complaints. [2]

(c) On a randomly chosen day, find the most likely number of complaints received. Justify your answer. [3]

The department store introduces a new policy to improve customer service. The number of complaints received per day now follows a Poisson distribution with mean λ .

On a randomly chosen day, the probability that there are no complaints is now 0.8.

(d) Find the value of λ . [2]

16EP11


ApTutorGroup.com

– 12 –



The vertical cross-section of a container is shown in the following diagram.

10

20

30

40

50

010 20–10–20

y

x

The curved sides of the cross-section are given by the equation y = 0.25x2 - 16 . The horizontal cross-sections are circular. The depth of the container is 48 cm .

(a) If the container is filled with water to a depth of h cm, show that the volume, V cm3, of

the water is given by V h h= +

4

216

2

π . [3]

(This question continues on the following page)

16EP12


ApTutorGroup.com

– 13 –

Turn over


(Question 10 continued)

The container, initially full of water, begins leaking from a small hole at a rate given by ddVt

hh

= −+

25016( )π

where t is measured in seconds.

(b) (i) Show that ddht

hh

= −+

2504 162 2( )π

.

(ii) State ddth

and hence show that t h h h h=−

+ +

−4250

32 2562 3

212

12 d∫ π

.

(iii) Find, correct to the nearest minute, the time taken for the container to become empty. (60 seconds = 1 minute) [11]

Once empty, water is pumped back into the container at a rate of 8.5 cm3 s-1 . At the same

time, water continues leaking from the container at a rate of 250

163 1h

h( )+−cm s

π.

(c) Using an appropriate sketch graph, determine the depth at which the water ultimately stabilizes in the container. [3]

16EP13


ApTutorGroup.com

– 14 –



In triangle ABC,

3 sin B + 4 cos C = 6 and

4 sin C + 3 cos B = 1 .

(a) Show that sin ( )B C+ =12

. [6]

Robert conjectures that ˆCAB can have two possible values.

(b) Show that Robert’s conjecture is incorrect by proving that ˆCAB has only one possible value. [5]

16EP14


ApTutorGroup.com

Please do not write on this page.

Answers written on this page will not be marked.

16EP15

ApTutorGroup.com

ApTutorGroup.com

Please do not write on this page.

Answers written on this page will not be marked.

16EP16

ApTutorGroup.com

ApTutorGroup.com

SPEC/5/MATHL/HP2/ENG/TZ0/XX/M

16 pages

Markscheme


Mathematics

Higher level

Paper 2

ApTutorGroup.com

ApTutorGroup.com



Abbreviations M Marks awarded for attempting to use a valid Method; working must be seen. (M) Marks awarded for Method; may be implied by correct subsequent working.

A Marks awarded for an Answer or for Accuracy; often dependent on preceding M marks.



Mark according to RM™ Assessor instructions and the document “Mathematics HL: Guidance for e-marking May 2016”. It is essential that you read this document before you start marking. In particular, please note the following. Marks must be recorded using the annotation stamps. Please check that you are entering marks for the right question. If a part is completely correct, (and gains all the ‘must be seen’ marks), use the ticks with





Where M and A marks are noted on the same line, for example, M1A1, this usually means M1 for an attempt to use an appropriate method (for example, substitution into a formula) and A1 for using the correct values.

Where the markscheme specifies (M2), N3, etc, do not split the marks.

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ApTutorGroup.com


Once a correct answer to a question or part-question is seen, ignore further correct working. However, if further working indicates a lack of mathematical understanding do not award the final A1. An exception to this may be in numerical answers, where a correct exact value is followed by an incorrect decimal. However, if the incorrect decimal is carried through to a subsequent part, and correct FT working shown, award FT marks as appropriate but do not award the final A1 in that part.

Examples




2. 1sin 4






4 Implied marks Implied marks appear in brackets, for example, (M1), and can only be awarded if correct work is

seen or if implied in subsequent working. Normally the correct work is seen or implied in the next line. Marks without brackets can only be awarded for work that is seen.






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ApTutorGroup.com


6 Misread






8 Alternative methods

Candidates will sometimes use methods other than those in the markscheme. Unless the question specifies a method, other correct methods should be marked in line with the markscheme. If in doubt, contact your team leader for advice.



9 Alternative forms

Unless the question specifies otherwise, accept equivalent forms.






10 Accuracy of Answers Candidates should NO LONGER be penalized for an accuracy error (AP). If the level of accuracy is specified in the question, a mark will be allocated for giving the answer to


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ApTutorGroup.com


11 Crossed out work If a candidate has drawn a line through work on their examination script, or in some other way crossed out their work, do not award any marks for that work.

12 Calculators

A GDC is required for paper 2, but calculators with symbolic manipulation features (for example, TI-89) are not allowed.

Calculator notation The Mathematics HL guide says: Students must always use correct mathematical notation, not calculator notation.

Do not accept final answers written using calculator notation. However, do not penalize the use of calculator notation in the working.



14 Candidate work

Candidates are meant to write their answers to Section A on the question paper (QP), and Section B on answer booklets. Sometimes, they need more room for Section A, and use the booklet (and often comment to this effect on the QP), or write outside the box. This work should be marked.

The instructions tell candidates not to write on Section B of the QP. Thus they may well have done some rough work here which they assume will be ignored. If they have solutions on the answer booklets, there is no need to look at the QP. However, if there are whole questions or whole part solutions missing on answer booklets, please check to make sure that they are not on the QP, and if they are, mark those whole questions or whole part solutions that have not been written on answer booklets.

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ApTutorGroup.com


Section A

1. 1

4

2

1

n and 2

1

3

3

n (A1)(A1)

use of 1 2

1 2

cos n n

n n (M1)

7 7

cos21 19 399

(A1)(A1)

Note: Award A1 for a correct numerator and A1 for a correct denominator.

69 A1 Note: Award A1 for 111 .

[6 marks] 2. (a) P ( ) 0.99 ( P ( ) 0.01)X x X x (M1) 54.6 (cm)x A1 [2 marks] (b) P (60.15 60.25)X (M1)(A1)(A1) 0.0166 A1 [4 marks] Total [6 marks]

3. (a) attempting to find a normal to eg

3 8

2 11

2 6

(M1)

3 8 2

2 11 17 2

2 6 1

(A1)

2 1 2

2 5 2

1 12 1

r M1

2 2 4x y z (or equivalent) A1 [4 marks]

continued…

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(b) 3

4 2

: 0 2 ,

8 1

l t t

r (A1)

attempting to solve

4 2 2

2 2 4

8 1

ttt

for t ie 9 16 4t for t M1

4

3t A1

4 8 20, ,

3 3 3

A1

[4 marks] Total [8 marks] 4. using ( ) 7p a to obtain 3 23 5 7 0a a a M1A1 2( 1) (3 2 7) 0a a a (M1)(A1)

Note: Award M1 for a cubic graph with correct shape and A1 for clearly showing that the above cubic crosses the horizontal axis at ( 1, 0) only.

1a A1 EITHER

showing that 23 2 7 0a a has no real (two complex) solutions for a R1 OR

showing that 3 23 5 7 0a a a has one real (and two complex) solutions for a R1

Note: Award R1 for solutions that make specific reference to an appropriate graph.

Total [6 marks]

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5. (a) using 32

1 2

uuru u

to form 2

6 2

a d aa d a d

(M1)

2( 6 ) ( 2 )a a d a d A1

2 (2 ) 0d d a (or equivalent) A1

since 02

ad d AG

[3 marks]

(b) substituting 2

ad into 6 3a d and solving for a and d (M1)

3

4a and

3

8d (A1)

1

2r A1

13 1

23 32 ( 1) 200

12 4 8 12

n

n n

(A1)

attempting to solve for n (M1) 31.68n

so the least value of n is 32 A1 [6 marks] Total [9 marks]

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6. (a) 3 0 6(s)2

t t (M1)A1

[2 marks]

Note: Award A0 if either 0.236t or 4.24t or both are stated with 6t . (b) let d be the distance travelled before coming to rest

4 6

2

0 4

5 ( 2) d 3 d2

td t t t (M1)(A1)

Note: Award M1 for two correct integrals even if the integration limits are incorrect.

The second integral can be specified as the area of a triangle.

47

( 15.7) (m)3

d (A1)

attempting to solve 6

473 d

2 3

T t t (or equivalent) for T M1

13.9 (s)T A1 [5 marks] Total [7 marks]

7. (a) each triangle has area 21 2sin

8x

n

1use of sin2

ab C (M1)

there are n triangles so 21 2sin

8A nx

n

A1

2

2

1 2π4 sin

8π

nxnC

x

A1

so 2π

sin2π

nCn

AG

[3 marks]

continued…

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(b) attempting to find the least value of n such that 2π

sin 0.992π

nn (M1)

26n A1

attempting to find the least value of n such that

2πsin

0.99π

π 1 cos

nn

n

(M1)

21n and so a regular polygon with 21 sides A1

Note: Award (M0)A0(M1)A1 if 2π

sin 0.992π

nn is not considered and

2πsin

0.99π

π 1 cos

nn

n

is

correctly considered. Award (M1)A1(M0)A0 for 26n .

[4 marks]

(c) EITHER

for even and odd values of n, the value of C seems to increase towards the limiting value of the circle ( 1)C ie as n increases, the polygonal regions get closer and closer to the enclosing circular region R1

OR

the differences between the odd and even values of n illustrate that this measure of compactness is not a good one R1

[1 mark] Total [8 marks]

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Section B

8. (a) use of 1

sin2

A qr to obtain 21( 2) (5 ) sin 30

2A x x M1

21( 2) (25 10 )

4x x x A1

3 21( 8 5 50)

4A x x x AG

[2 marks]

(b) (i) 2d 1 13 16 5 (3 1) ( 5)

d 4 4

A x x x xx

A1

(ii) METHOD 1 EITHER

2

d 1 1 13 16 5 0

d 4 3 3

Ax

M1A1

OR

d 1 1 1

3 1 5 0d 4 3 3

Ax

M1A1

THEN

so d

0d

Ax when

1

3x AG

METHOD 2

solving d

0d

Ax for x M1

1

2 53

x x A1

so d

0d

Ax when

1

3x AG

METHOD 3

a correct graph of d

d

Ax

versus x M1

the graph clearly showing that d

0d

Ax when

1

3x A1

so d

0d

Ax when

1

3x AG

[3 marks]

continued…

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(c) (i) 2

2

d 1(3 8)

d 2

A xx

A1

for 1

3x ,

2

2

d3.5 ( 0)

d

Ax

R1

so 1

3x gives the maximum area of triangle PQR AG

(ii) 2max

343( 12.7) (cm )

27A A1

(iii) 7

PQ3

(cm) and 2

14PR

3

(cm) (A1)

2 4 2

2 7 14 7 14QR 2 cos30

3 3 3 3

(M1)(A1)

391.702 QR 19.8(cm) A1 [7 marks] Total [12 marks] 9. (a) (i) 0.6P ( 0) 0.549 eX A1

(ii) P ( 3) 1 P ( 2)X X (M1)

P ( 3) 0.0231X A1 [3 marks] (b) EITHER

using Po(3)Y (M1) OR

using 5(0.549) (M1) THEN

3P ( 0) 0.0498 eY A1

[2 marks]

continued…

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Question 9 continued (c) P ( 0)X (most likely number of complaints received is zero) A1 EITHER

calculating P ( 0) 0.549X and P ( 1) 0.329X M1A1 OR

sketching an appropriate (discrete) graph of P ( )X x against x M1A1 OR

finding 0.6P ( 0)X e and stating that P ( 0) 0.5X M1A1 OR

using P ( ) P ( 1)X x X xx

where 1 M1A1

[3 marks] (d) P( 0) 0.8 0.8X e (A1)

5 40.223 ln , ln

4 5

A1


10. (a) attempting to use 2π db

aV x y (M1)

attempting to express 2x in terms of y ie 2 4( 16)x y (M1)

for y h , 0

4π 16dh

V y y A1

2

4π 162

hV h

AG

[3 marks]

continued…

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Question 10 continued (b) (i) METHOD 1

d d d

d d d

h h Vt V t (M1)

d

4π ( 16)d

V hh (A1)

d 1 250

d 4π ( 16) π ( 16)

h ht h h

M1A1

Note: Award M1 for substitution into d d d

d d d

h h Vt V t .

2 2

d 250

d 4π ( 16)

h ht h

AG

METHOD 2

d d

4π( 16)d d

V hht t (implicit differentiation) (M1)

250 d

4π( 16)π ( 16) d

h hhh t

(or equivalent) A1

d 1 250

d 4π ( 16) π ( 16)

h ht h h

M1A1

2 2

d 250

d 4π ( 16)

h ht h

AG

(ii) 2 2d 4π ( 16)

d 250

t hh h

A1

2 24π ( 16)

d250

ht hh

(M1)

2 24π ( 32 256)

d250

h ht hh

A1

3 1 122 2 2

4π32 256 d

250t h h h h

AG

continued…

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Question 10 continued (iii) METHOD 1

0 3 1 1

2 2 2

48

32 256 dt h h h h

(M1)(A1)

2688.756...t (s) (A1) 45 minutes (correct to the nearest minute) A1 METHOD 2

5 3 122 2 2

4π 2 64512

250 5 3t h h h c

A1

5 3 122 2 2

4π 2 640, 48 2688.756... 48 48 512 48

250 5 3t h c c

(M1)

h 0, 5 3 122 2 2

4π 2 642688.756... 48 48 512 48

250 5 3t t

(s) (A1)

45 minutes (correct to the nearest minute) A1 [11 marks] (c) EITHER

the depth stabilizes when d

0d

Vt ie

2508.5 0

π ( 16)

hh

R1

attempting to solve 250

8.5 0π ( 16)

hh

for h (M1)

OR

the depth stabilizes when d

0d

ht ie

1 2508.5 0

4π ( 16) π ( 16)

hh h

R1

attempting to solve 1 250

8.5 04π ( 16) π ( 16)

hh h

for h (M1)

THEN

5.06h (cm) A1 [3 marks] Total [17 marks]

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11. (a) squaring both equations M1

2 29sin 24sin cos 16cos 36B B C C (A1)

2 29cos 24cos sin 16sin 1B B C C (A1)

adding the equations and using 2 2cos sin 1 to obtain 9 24(sin cos cos sin ) 16 37B C B C M1

24 (sin cos cos sin ) 12B C B C A1

24sin ( ) 12B C (A1)

1

sin ( )2

B C AG

[6 marks] (b) sin sin 180 ( )A B C so sin sin ( )A B C R1

1 1

sin ( ) sin2 2

B C A A1

30A or 150A A1

Note: Award R1A1A1 for obtaining 30B C or 150B C . if 150A , then 30B R1

for example, 33sin 4cos 4 6

2B C

, ie a contradiction R1

only one possible value ( 30 )A AG [5 marks] Total [11 marks]

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SPEC/5/MATHL/HP3/ENG/TZ0/DM

MathematicsHigher levelPaper 3 – discrete mathematics



yy Do not open this examination paper until instructed to do so.yy Answer all the questions.yy Unless otherwise stated in the question, all numerical answers should be given exactly or

correct to three significant figures.yy A graphic display calculator is required for this paper.yy A clean copy of the mathematics HL and further mathematics HL formula booklet is


1 hour


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– 2 –

Please start each question on a new page. Full marks are not necessarily awarded for a correct answer with no working. Answers must be supported by working and/or explanations. In particular, solutions found from a graphic display calculator should be supported by suitable working. For example, if graphs are used to find a solution, you should sketch these as part of your answer. Where an answer is incorrect, some marks may be given for a correct method, provided this is shown by written working. You are therefore advised to show all working.


Let f (n) = n5 - n , n ∈ + .

(a) Find the values of f (3) and f (4) . [1]

(b) Use the Euclidean algorithm to find gcd( f (3) , f (4)) . [2]

(c) Use Fermat’s Little Theorem to explain why f (n) is always exactly divisible by 5. [1]

(d) By factorizing f (n) explain why it is always exactly divisible by 6. [4]


(a) Use the pigeon-hole principle to prove that for any simple graph that has two or more vertices and in which every vertex is connected to at least one other vertex, there must be at least two vertices with the same degree. [4]

Seventeen people attend a meeting.

(b) Each person shakes hands with at least one other person and no-one shakes hands with the same person more than once. Use the result from part (a) to show that there must be at least two people who shake hands with the same number of people. [4]

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– 3 –

Turn over


The following graph represents the cost in dollars of travelling by bus between 10 towns in a particular province.

11 9

61311

5

87

718

10

11 95

E

F

G

H

J

IB

A

C

D

5 4

(a) Use Dijkstra’s algorithm to find the cheapest route between A and J , and state its cost. [7]

For the remainder of the question you may find the cheapest route between any two towns by inspection.

It is given that the total cost of travelling on all the roads without repeating any is $139 . A tourist decides to go over all the roads at least once, starting and finishing at town A .

(b) Find the lowest possible cost of his journey, stating clearly which roads need to be travelled more than once. You must fully justify your answer. [6]


(a) Solve, by any method, the following system of linear congruences

x ≡ 9(mod 11)x ≡ 1(mod 5). [3]

(b) Find the remainder when 4182 is divided by 11 . [4]

(c) Using your answers to parts (a) and (b) find the remainder when 4182 is divided by 55 . [3]


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– 4 –


Andy and Roger are playing tennis with the rule that if one of them wins a game when serving then he carries on serving, and if he loses then the other player takes over the serve.

The probability Andy wins a game when serving is 12

and the probability he wins a game

when not serving is 14

. Andy serves in the first game. Let un denote the probability that

Andy wins the nth game.

(a) State the value of u1 . [1]

(b) Show that un satisfies the recurrence relation

u un n= +−

14

141 . [4]

(c) Solve this recurrence relation to find the probability that Andy wins the nth game. [6]


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SPEC/5/MATHL/HP3/ENG/TZ0/DM/M

9 pages

Markscheme


Discrete mathematics

Higher level

Paper 3

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Abbreviations M Marks awarded for attempting to use a valid Method; working must be seen. (M) Marks awarded for Method; may be implied by correct subsequent working. A Marks awarded for an Answer or for Accuracy; often dependent on preceding M marks.



Mark according to RM™ Assessor instructions and the document “Mathematics HL: Guidance for e-marking May 2016”. It is essential that you read this document before you start marking. In particular, please note the following: Marks must be recorded using the annotation stamps. Please check that you are entering

marks for the right question. If a part is completely correct, (and gains all the “must be seen” marks), use the ticks with





Where M and A marks are noted on the same line, eg M1A1, this usually means M1 for an attempt to use an appropriate method (eg substitution into a formula) and A1 for using the correct values.

Where the markscheme specifies (M2), N3, etc., do not split the marks.

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Once a correct answer to a question or part-question is seen, ignore further correct working. However, if further working indicates a lack of mathematical understanding do not award the final A1. An exception to this may be in numerical answers, where a correct exact value is followed by an incorrect decimal. However, if the incorrect decimal is carried through to a subsequent part, and correct FT working shown, award FT marks as appropriate but do not award the final A1 in that part. Examples




2. 1 sin 44

x sin x Do not award the final A1





4 Implied marks Implied marks appear in brackets eg (M1), and can only be awarded if correct work is seen or if

implied in subsequent working. Normally the correct work is seen or implied in the next line. Marks without brackets can only be awarded for work that is seen.






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6 Misread






8 Alternative methods Candidates will sometimes use methods other than those in the markscheme. Unless the question specifies a method, other correct methods should be marked in line with the markscheme. If in doubt, contact your team leader for advice.



9 Alternative forms Unless the question specifies otherwise, accept equivalent forms.









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11 Crossed out work


12 Calculators

A GDC is required for paper 3, but calculators with symbolic manipulation features (eg TI-89) are not allowed.

Calculator notation The mathematics HL guide says: Students must always use correct mathematical notation, not calculator notation. Do not accept final answers written using calculator notation. However, do not penalize the use of calculator notation in the working.



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1. (a) 240, 1020 A1

Note: Award A2 for three correct answers, A1 for two correct answers. [1 mark] (b) 1020 240 4 60 (M1) 240 60 4 gcd (1020 , 240) 60 A1 [2 marks] Note: Must be done by Euclid’s algorithm.

(c) by Fermat’s little theorem with 5p

5 (mod 5)n n A1

so 5 divides ( )f n [1 mark] (d) 2 2 2( ) ( 1)( 1) ( 1)( 1)( 1)f n n n n n n n n (A1)A1

1, , 1n n n are consecutive integers and so contain a multiple of 2 and 3 R1R1

Note: Award R1 for justification of 2 and R1 for justification of 3.

and therefore ( )f n is a multiple of 6 AG [4 marks] Total [8 marks] 2. (a) let there be v vertices in the graph; because the graph is simple the degree

of each vertex is 1v A1 the degree of each vertex is 1 A1 there are therefore 1v possible values for the degree of each vertex A1 given there are v vertices by the pigeon-hole principle there must be at least

two with the same degree R1 [4 marks] (b) consider a graph in which the people at the meeting are represented by the

vertices and two vertices are connected if the two people shake hands M1 the graph is simple as no-one shakes hands with the same person more than

once (nor can someone shake hands with themselves) A1 every vertex is connected to at least one other vertex as everyone shakes at

least one hand A1 the degree of each vertex is the number of handshakes so by the proof

above there must be at least two who shake the same number of hands R1 [4 marks] Total [8 marks]

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3. (a)

M1A1A1A1

(M1 for an attempt at Dijkstra’s) (A1 for value of D 17 ) (A1 for value of H 22 ) (A1 for value of G 24 ) route is ABDHJ (M1)A1 cost is $27 A1 Note: Accept other layouts.

[7 marks] (b) there are 4 odd vertices A, D, F and J A1 these can be joined up in 3 ways with the following extra costs AD and FJ 17 13 30 AF and DJ 23 10 33 AJ and DF 27 12 39 M1A1A1

Notes: Award M1 for an attempt to find different routes. Award A1A1 for correct values for all three costs A1 for one correct.

need to repeat AB, BD, FG and GJ A1 total cost is 139 30 $169 A1 [6 marks] Total [13 marks]

A B C D E F G H I J

A 0 10 11 18

B 10 17 21 23

C 11

D 17 24 22

E 21 30

H 22 27

F 23 29

G 24 28

J 27

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4. (a) METHOD 1

listing 9 , 20 , 31, and 1, 6 ,11,16, 21, 26, 31, M1 one solution is 31 (A1) by the Chinese remainder theorem the full solution is 31(mod 55)x A1 N2 METHOD 2

9(mod11) 9 11x x t M1

9 11 1(mod5)t 2(mod5)t A1 2 5t s

9 11(2 5 )x s

31 55 31(mod55)x s x A1

Note: Accept other methods eg formula, Diophantine equation. Note: Accept other equivalent answers eg –79(mod55).

[3 marks]

(b) 82 8241 8 (mod11)

by Fermat’s little theorem 108 1(mod11) (or 1041 1(mod11) ) M1

82 28 8 (mod11) M1

9(mod11) (A1)

remainder is 9 A1 [4 marks] Note: Accept simplifications done without Fermat.

(c) 82 8241 1 1(mod5) A1

so 8241 has a remainder 1 when divided by 5 and a remainder 9 when divided by 11 R1

hence by part (a) the remainder is 31 A1 [3 marks] Total [10 marks]

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5. (a) 12

A1

[1 mark] (b) Andy could win the nth game by winning the 1thn and then winning

the nth game or by losing the 1thn and then winning the nth (M1)

1 11 1 (1 )2 4n n nu u u A1A1M1

Note: Award A1 for each term and M1 for addition of two probabilities.

11 14 4n nu u AG

[4 marks]

(c) general solution is 1 ( )4

n

nu A p n

(M1)

for a particular solution try ( )p n b (M1)

1 14 4

b b (A1)

13

b

hence 1 14 3

n

nu A

(A1)

using 112

u M1

1 1 1 22 4 3 3

A A

hence 2 1 13 4 3

n

nu

A1

Note: Accept other valid methods.


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SPEC/5/MATHL/HP3/ENG/TZ0/SE

MathematicsHigher levelPaper 3 – calculus






1 hour


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– 2 –



(a) Use the integral test to determine the convergence or divergence of

0.51

1n n

∞

=∑ . [3]

(b) Let S xn

n

nn

=+×=

( ).

12 0 5

1

∞

∑ .

(i) Use the ratio test to show that S is convergent for -3 < x < 1 .

(ii) Hence find the interval of convergence for S . [11]


(a) Use an integrating factor to show that the general solution for ddxtxt t

t− = − >2 0, is

x = 2 + ct , where c is a constant. [4]

The weight in kilograms of a dog, t weeks after being bought from a pet shop, can be modelled by the following function

w tct t

tt

( ) =− >

2 0 5

16 35 5

+ ≤ ≤.

(b) Given that w(t) is continuous, find the value of c . [2]

(c) Write down an upper bound for the weight of the dog. [1]

(d) Prove from first principles that w(t) is differentiable at t = 5 . [6]

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– 3 –


Consider the differential equation ddyx

f x y= ( , ) where f (x , y) = y - 2x .

(a) Sketch, on one diagram, the four isoclines corresponding to f (x , y) = k where k takes the values -1 , -0.5 , 0 and 1. Indicate clearly where each isocline crosses the y-axis. [2]

A curve, C, passes through the point (0 , 1) and satisfies the differential equation above.

(b) Sketch C on your diagram. [3]

(c) State a particular relationship between the isocline f (x , y) = - 0.5 and the curve C, at their point of intersection. [1]

(d) Use Euler’s method with a step interval of 0.1 to find an approximate value for y on C, when x = 0.5 . [4]


In this question you may assume that arctan x is continuous and differentiable for x ∈ .

(a) Consider the infinite geometric series

1 - x2 + x4 - x6 + … | x | < 1 .

Show that the sum of the series is 1

1 2+ x. [1]

(b) Hence show that an expansion of arctan x is arctan x x x x x= − + − +

3 5 7

3 5 7… [4]

(c) f is a continuous function defined on [a , b] and differentiable on ]a , b[ with f '(x) > 0 on ]a , b[ .

Use the mean value theorem to prove that for any x , y ∈ [a , b] , if y > x then f (y) > f (x) . [4]

(d) (i) Given g(x) = x - arctan x , prove that g'(x) > 0 , for x > 0 .

(ii) Use the result from part (c) to prove that arctan x < x , for x > 0 . [4]

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SPEC/5/MATHL/HP3/ENG/TZ0/SE/M

10 pages

Markscheme


Calculus

Higher level

Paper 3

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– 2 – SPEC/5/MATHL/HP3/ENG/TZ0/SE/M













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2. 1sin 4













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6 Misread



the final answer(s). 7 Discretionary marks (d) An examiner uses discretion to award a mark on the rare occasions when the markscheme does














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11 Crossed out work


12 Calculators





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1. (a) 0.5

1dx x

M1

0.5

1lim 2

H

Hx

A1

Note: Accept 0.5

12x

.

this is not finite so series is divergent R1

Notes: Accept equivalent eg , or “limit does not exist”. If lower limit is not equal to 1 award M0A0, but the R1 can still be awarded if the final reasoning is correct.

[3 marks] (b) (i) applying the ratio test M1

1 0.5

1 0.5

( 1) 2lim

2 ( 1) ( 1)

n n

n nn

x nn x

A1

0.5

0.5

( 1) ( 1)lim

22( 1)n

x n xn

A1

Note: Do not penalize the absence of limits and modulus signs.

converges if 1 ( 1)

1 1 12 2

x x M1

3 1x A1

Note: Accept 2 1 2x . (ii) considering end points M1

when 3x , series is 0.5

1

( 1)n

n n

A1

0.5

1

n is a decreasing sequence with limit zero, R1

so series converges by alternating series test R1

when 1x , series is 0.51

1

n n

which diverges by part (a) or p-series A1

Note: This A1 is for both the reasoning and the statement it diverges.

interval of convergence is 3 1x A1 [11 marks] Total [14 marks]

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2. (a) integrating factor 1

d ln 1e e

t tt

t

M1A1

2

2 2d

x t ct t t A1A1

Note: Award A1 for xt

and A1 for 2 ct .

2x ct AG [4 marks] (b) given continuity at 5x

35 75 2 16

5 5c c M1A1

[2 marks] (c) any value ≥ 16 A1 Note: Accept values less than 16 if fully justified by reference to the maximum age for a dog.

[1 mark]

(d) 0

7 7(5 ) 2 (5) 2 75 5lim

5h

h

h

M1A1

0

35 3516 16

5 5limh

hh

0

357

5limh

hh

M1

0 0

35 35 77 7(5 )

lim lim5 5h h

hh

h h

M1A1

both limits equal so differentiable at 5t R1AG

Notes: The limits 5t could also be used.

For each value of 7

5 obtained by standard differentiation award A1.

To gain the other [4 marks] a rigorous explanation must be given on how you can get from the left and right hand derivatives to the derivative.

Notes: If the candidate works with t and then substitutes t 5 at the end award as follows.

First M1 for using formula with t in the linear case, A1 for 7

5.

Award next 2 method marks even if t 5 not substituted, A1 for 7

5.


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3. (a) and (b)

(a) A1 for 4 parallel straight lines with a positive gradient A1 A1 for correct y intercepts A1 [2 marks] (b) A1 for passing through (0 ,1) with positive gradient less than 2 A1 for stationary point on 2y x A1 for negative gradient on both of the other 2 isoclines A1A1A1 [3 marks] (c) the isocline is perpendicular to C R1 [1 mark] (d) 1 0.1( 2 )( 1.1 0.2 )n n n n n ny y y x y x (M1)(A1)

Note: Also award M1A1 if no formula seen but 2y is correct.

0 1 2 3 41, 1.1, 1.19, 1.269, 1.3359y y y y y (M1)

5 1.39y to 3 sf A1

Note: M1 is for repeated use of their formula, with steps of 0.1.

Note: Accept 1.39 or 1.4 only. [4 marks] Total [10 marks]

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4. (a) 2

2

1,

1r x S

x

A1AG

[1 mark]

(b) 2 4 62

11

1x x x

x

EITHER

2 4 62

1d 1 d

1x x x x x

x

M1

3 5 7

arctan3 5 7

x x xx c x A1

Note: Do not penalize the absence of c at this stage.

when 0x we have arctan 0 c hence 0c M1A1 OR

2 4 620 0

1d 1 d

1

x xt t t t t

t

M1A1A1

Notes: Allow x as the variable as well as the limit. M1 for knowing to integrate, A1 for each of the limits.

3 5 7

00

arctan3 5 7

xx t t tt t

A1

hence 3 5 7

arctan3 5 7

x x xx x AG

[4 marks] (c) applying the MVT to the function f on the interval [ , ]x y M1

( ) ( )

( )f y f x f c

y x

(for some ,c x y ) A1

( ) ( )

0f y f x

y x

(as ( ) 0f c ) R1

( ) ( ) 0f y f x as y x R1 ( ) ( )f y f x AG [4 marks] Note: If they use x rather than c they should be awarded M1A0R0, but could get the next R1.

continued…

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(d) (i) ( ) arctang x x x 2

1( ) 1

1g x

x

A1

this is greater than zero because 2

11

1 x

R1

so ( ) 0g x AG (ii) (g is a continuous function defined on [0 , ]b and differentiable on

0, b with ( ) 0g x on 0, b for all b )

(if 0 ,x b then) from part (c) ( ) (0)g x g M1

arctan 0 arctanx x x x M1

(as b can take any positive value it is true for all x 0) AG [4 marks] Total [13 marks]

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SPEC/5/MATHL/HP3/ENG/TZ0/SG

MathematicsHigher levelPaper 3 – sets, relations and groups






1 hour


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– 2 –



A group with the binary operation of multiplication modulo 15 is shown in the following Cayley table.

×15 1 2 4 7 8 11 13 14

1 1 2 4 7 8 11 13 14

2 2 4 8 14 1 7 11 13

4 4 8 1 13 2 14 7 11

7 7 14 13 4 11 2 1 8

8 8 1 2 11 4 13 14 7

11 11 7 14 2 13 a b c

13 13 11 7 1 14 d e f

14 14 13 11 8 7 g h i

(a) Find the values represented by each of the letters in the table. [3]

(b) Find the order of each of the elements of the group. [3]

(c) Write down the three sets that form subgroups of order 2. [2]

(d) Find the three sets that form subgroups of order 4. [4]

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– 3 –

Turn over


Define f : \ {0.5} → by f x xx

( ) =+−

4 12 1

.

(a) Prove that f is an injection. [4]

(b) Prove that f is not a surjection. [4]


Consider the set A consisting of all the permutations of the integers 1 , 2 , 3 , 4 , 5 .

(a) Two members of A are given by p = (1 2 5) and q = (1 3)(2 5) . Find the single permutation which is equivalent to q p . [3]

(b) State a permutation belonging to A of order 6 . [2]

(c) Let P = {all permutations in A where exactly two integers change position},

and Q = {all permutations in A where the integer 1 changes position}.

(i) List all the elements in P ∩ Q .

(ii) Find n(P ∩ Q' ) . [4]


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The group {G , *} has identity eG and the group {H , } has identity eH . A homomorphism f is such that f : G → H . It is given that f (eG ) = eH .

(a) Prove that for all a ∈ G , f (a-1) = ( f (a))-1 . [4]

Let {H , } be the cyclic group of order seven, and let p be a generator. Let x ∈ G such that f (x) = p2 .

(b) Find f (x-1) . [2]

(c) Given that f (x * y) = p , find f ( y) . [4]


{G , *} is a group with identity element e . Let a , b ∈ G .

(a) Verify that the inverse of a * b-1 is equal to b * a-1 . [3]

Let {H , *} be a subgroup of {G , *} . Let R be a relation defined on G by

aRb ⇔ a * b-1 ∈ H .

(b) Prove that R is an equivalence relation, indicating clearly whenever you are using one of the four properties required of a group. [8]


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SPEC/5/MATHL/HP3/ENG/TZ0/SG/M

9 pages

Markscheme


Sets, relations and groups

Higher level

Paper 3

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– 2 – SPEC/5/MATHL/HP3/ENG/TZ0/SG/M













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2. 1 sin 44

x sin x Do not award the final A1












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6 Misread


















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11 Crossed out work


12 Calculators





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1. (a) 1, 8, 4,a b c 8, 4, 2,d e f 4, 2, 1g h i A3

Note: Award A3 for 9 correct answers, A2 for 6 or more, and A1 for 3 or more. [3 marks]

(b) Elements Order 1 1 4, 11, 14 2 2, 7, 8, 13 4 A3

Note: Award A3 for 8 correct answers, A2 for 6 or more, and A1 for 4 or more.

[3 marks] (c) {1, 4}, {1, 11}, {1, 14} A1A1

Note: Award A1 for 1 correct answer and A2 for all 3 (and no extras). [2 marks] (d) {1, 2, 4, 8}, {1, 4, 7, 13}, A1A1 {1, 4, 11, 14} A2 [4 marks] Total [12 marks]

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2. (a) METHOD 1

4 1 4 1( ) ( )2 1 2 1x yf x f yx y

M1A1

for attempting to cross multiply and simplify M1 (4 1)(2 1) (2 1)(4 1)x y x y

8 2 4 1 8 2 4 1xy y x xy x y 6 6y x x y A1 hence an injection AG [4 marks] METHOD 2

2 24(2 1) 2(4 1) 6( )

(2 1) (2 1)x xf x

x x

M1A1

0 (for all 0.5x ) R1 therefore the function is decreasing on either side of the discontinuity and

( ) 2f x for 0.5x and 2x for ( ) 0.5f x R1 hence an injection AG

Note: If a correct graph of the function is shown, and the candidate states this is decreasing in each part (or horizontal line test) and hence an injection, award M1A1R1.

[4 marks] (b) METHOD 1

attempt to solve 4 12 1xyx

M1

(2 1) 4 1 2 4 1y x x xy y x A1

12 4 12 4

yxy x y xy

A1

no value for 2y R1 hence not a surjection AG [4 marks] METHOD 2

consider 2y A1

attempt to solve 4 122 1xx

M1

4 2 4 1x x A1 which has no solution R1 hence not a surjection AG

Note: If a correct graph of the function is shown, and the candidate states that because there is a horizontal asymptote at 2y then the function is not a surjection, award M1R1.


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3. (a) (1 3) (2 5) (1 2 5)q p (M1)

(1 5 3) M1A1

Note: M1 for an answer consisting of disjoint cycles, A1 for (1 5 3).

Note: Allow 1 2 3 4 55 2 1 4 3

, allow (1 5 3)(2).

If done in the wrong order and obtained (1 3 2), award A2. [3 marks] (b) any permutation with 2 disjoint cycles one of length 2 and one of

length 3 eg (1 2) (3 4 5) M1A1

Notes: Award M1A0 for any permutation with 2 non-disjoint cycles one of length 2 and one of length 3. Accept non cycle notation.

[2 marks] (c) (i) (1, 2), (1, 3), (1, 4), (1, 5) M1A1 (ii) (2 3), (2 4), (2 5), (3 4), (3 5), (4 5) (M1) 6 A1

Note: Award M1 for at least one correct cycle. [4 marks] Total [9 marks] 4. (a) ( )G Hf e e 1

Hf a a e M1

f is a homomorphism so 1 1* ( ) Hf a a f a f a e M1A1

by definition 1( ) ( ) Hf a f a e so 11 ( )f a f a (by the left-cancellation law) R1

[4 marks]

(b) from (a) 1 1( )f x f x

hence 1 2 1 5( ) ( )f x p p M1A1 [2 marks] (c) ( * ) ( ) ( )f x y f x f y (homomorphism) (M1)

2 ( )p f y p A1

5( )f y p p (M1)

6p A1 [4 marks] Total [10 marks]

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5. (a) METHOD 1

1 1 1 1 1 1a b b a a b b a a e a a a e M1A1A1

Notes: M1 for multiplying, A1 for at least one of the next 3 expressions, A1 for e.

Allow 1 1 1 1 1 1* * * * * * * * *b a a b b a a b b e b b b e .

METHOD 2

1 11 1 1a b b a M1A1

1*b a

A1 [3 marks]

(b) 1*a a e H (as H is a subgroup) M1

so aRa and hence R is reflexive

1*aRb a b H . H is a subgroup so every element has an inverse in

H so 1*a b H R1

1*b a H bRa

M1

so R is symmetric

1 1, * , *aRb bRc a b H b c H

M1

as H is closed 1 1* * *a b b c H R1

and using associativity R1

1 1 1 1 1* * * * * * *a b b c a b b c a c H aRc A1

therefore R is transitive R is reflexive, symmetric and transitive A1

Note: Can be said separately at the end of each part. hence it is an equivalence relation AG [8 marks] Total [11 marks]

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SPEC/5/MATHL/HP3/ENG/TZ0/SP

MathematicsHigher levelPaper 3 – statistics and probability






1 hour


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– 2 –



A random variable X has probability density function

f x

x

x

x

x

( ) =

<

≤ <

≤ <

≥

0 0120 1

141 3

0 3

.

(a) Sketch the graph of y = f (x) . [1]

(b) Find the cumulative distribution function for X . [5]

(c) Find the upper quartile for X . [1]


Eric plays a game at a fairground in which he throws darts at a target. Each time he throws a dart, the probability of hitting the target is 0.2 . He is allowed to throw as many darts as he likes, but it costs him $1 a throw. If he hits the target a total of three times he wins $10 .

(a) Find the probability he has his third success of hitting the target on his sixth throw. [3]

(b) (i) Find the expected number of throws required for Eric to hit the target three times.

(ii) Write down his expected profit or loss if he plays until he wins the $10 . [3]

(c) If he has just $8 , find the probability he will lose all his money before he hits the target three times. [3]

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– 3 –

Turn over


(a) If X and Y are two random variables such that E(X ) = µX and E(Y ) = µY then Cov(X , Y ) = E((X - µX)(Y - µY)).Prove that if X and Y are independent then Cov(X , Y ) = 0 . [3]

(b) In a particular company, it is claimed that the distance travelled by employees to work is independent of their salary. To test this, 20 randomly selected employees are asked about the distance they travel to work and the size of their salaries. It is found that the product moment correlation coefficient, r , for the sample is - 0.35 .

You may assume that both salary and distance travelled to work follow normal distributions.

Perform a one-tailed test at the 5 % significance level to test whether or not the distance travelled to work and the salaries of the employees are independent. [8]


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– 4 –


If X is a random variable that follows a Poisson distribution with mean λ > 0 then the probability generating function of X is G(t) = eλ(t-1) .

(a) (i) Prove that E(X ) = λ .

(ii) Prove that Var(X ) = λ . [6]

Y is a random variable, independent of X , that also follows a Poisson distribution with mean λ .

(b) If S = 2X - Y find

(i) E(S ) ;

(ii) Var(S ) . [3]

Let T X Y= +

2 2.

(c) (i) Show that T is an unbiased estimator for λ .

(ii) Show that T is a more efficient unbiased estimator of λ than S . [3]

(d) Could either S or T model a Poisson distribution? Justify your answer. [1]


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– 5 –


Two species of plant, A and B, are identical in appearance though it is known that the mean length of leaves from a plant of species A is 5.2 cm , whereas the mean length of leaves from a plant of species B is 4.6 cm . Both lengths can be modelled by normal distributions with standard deviation 1.2 cm .

In order to test whether a particular plant is from species A or species B, 16 leaves are collected at random from the plant. The length, x , of each leaf is measured and the mean length evaluated. A one-tailed test of the sample mean, X , is then performed at the 5 % level, with the hypotheses: H0 : µ = 5.2 and H1 : µ < 5.2 .

(a) Find the critical region for this test. [3]

(b) Find the probability of a Type II error if the leaves are in fact from a plant of species B. [2]

It is now known that in the area in which the plant was found 90 % of all the plants are of species A and 10 % are of species B.

(c) Find the probability that X will fall within the critical region of the test. [2]

(d) If, having done the test, the sample mean is found to lie within the critical region, find the probability that the leaves came from a plant of species A. [3]


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SPEC/5/MATHL/HP3/ENG/TZ0/SP/M

10 pages

Markscheme


Statistics and probability

Higher level

Paper 3

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– 2 – SPEC/5/MATHL/HP3/ENG/TZ0/SP/M


Abbreviations M Marks awarded for attempting to use a valid Method; working must be seen. (M) Marks awarded for Method; may be implied by correct subsequent working. A Marks awarded for an Answer or for Accuracy; often dependent on preceding M marks. (A) Marks awarded for an Answer or for Accuracy; may be implied by correct subsequent working. R Marks awarded for clear Reasoning. N Marks awarded for correct answers if no working shown. AG Answer given in the question and so no marks are awarded.


Mark according to RM™ Assessor instructions and the document “Mathematics HL: Guidance for e-marking May 2016”. It is essential that you read this document before you start marking. In particular, please note the following:

Marks must be recorded using the annotation stamps. Please check that you are entering








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2. 1sin 4













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6 Misread


















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11 Crossed out work


12 Calculators





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1. (a)

A1

Note: Ignore open / closed endpoints and vertical lines.

Note: Award A1 for a correct graph with scales on both axes and a clear indication of the relevant values.

[1 mark]

(b) 0 0

0 12( )

11 3

4 41 3

xx x

F xx x

x

considering the areas in their sketch or using integration (M1) ( ) 0,F x 0x , ( ) 1,F x 3x A1

( ) ,2

xF x 0 1x A1

1( ) ,

4 4

xF x 1 3x A1A1

Note: Accept < for ≤ in all places and also > for ≥ first A1.

[5 marks] (c) 3Q 2 A1

[1 mark] Total [7 marks]

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2. (a) METHOD 1

let X be the number of throws until Eric hits the target three times NB(3, 0.2)X (M1)

3 35P ( 6) 0.8 0.2

2X

(A1)

0.04096128

3125

(exact) A1

METHOD 2

let X be the number of hits in five throws X is B(5, 0.2) (M1)

3 35P ( 2) 0.2 0.8

2X

(0.2048) (A1)

P(3rd hit on 6th throw) 2 35 1280.2 0.8 0.2 0.04096

2 3125

(exact) A1

[3 marks]

(b) (i) expected number of throws 315

0.2 (M1)A1

(ii) profit 10 15 $5 or loss $5 A1

[3 marks] (c) METHOD 1

let Y be the number of times the target is hit in 8 throws (M1)

P ( 2)Y (M1)

0.797 A1 METHOD 2

let the 3rd hit occur on the Yth throw Y is NB (3, 0.2) (M1)

P ( 8) 1 – P ( 8)Y Y (M1)

0.797 A1 [3 marks]

Total [9 marks]

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3. (a) METHOD 1

Cov( , ) E ( ) ( )X YX Y X Y

E Y X X YXY X Y (M1)

E( ) E( ) E( )Y X X YXY X Y

E( ) X YXY A1

as X and Y are independent E( ) X YXY R1

Cov( , ) 0X Y AG METHOD 2

Cov( , ) E ( ) ( )X YX Y X Y

E Ex yX Y (M1)

since ,X Y are independent R1

( – )( – )x x y y A1

0 AG [3 marks] (b) H0 : 0 H1 : 0 A1

Note: The hypotheses must be expressed in terms of .

test statistic 2

20 20.35

1 0.35testt

(M1)(A1)

1.585 (A1) degrees of freedom 18 (A1) EITHER

p-value 0.0652 A1 this is greater than 0.05 M1 OR

5 %(18) 1.73t A1

this is less than –1.59 M1 THEN

hence accept 0H or reject 1H or equivalent or contextual equivalent R1

Note: Allow follow through for the final R1 mark. [8 marks] Total [11 marks]

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4. (a) (i) ( 1)( ) tG t e A1 E ( ) (1)X G M1

AG (ii) 2 ( 1)( ) tG t e M1

2(1)G (A1)

2Var ( ) (1) (1) (1)X G G G (M1)

2 2 A1 AG [6 marks] (b) (i) E ( ) 2S A1 (ii) Var ( ) 4 5S (A1)A1

Note: First A1 can be awarded for either 4 or + . [3 marks]

(c) (i) E ( )2 2

T (so T is an unbiased estimator) A1

(ii) 1 1 1Var ( )

4 4 2T A1

this is less than Var ( )S , therefore T is the more efficient estimator R1AG

Note: Follow through their variances from (b)(ii) and(c)(ii).

[3 marks] (d) no, mean does not equal the variance R1 [1 mark] Total [13 marks]

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5. (a) 21.2

5.2,16

X N

(M1)

critical value is 1.25.2 1.64485 4.70654

4 (A1)

critical region is , 4.71 A1

Note: Allow follow through for the final A1 from their critical value. [3 marks]

Note: Follow through previous values in (b), (c) and (d).

(b) type II error probability 21, 2

P 4.70654 is 4.6,16

X X N

(M1)

0.361 A1 [2 marks] (c) 0.9 0.05 0.1 (1 0.361 ) 0.108875997 0.109 M1A1

Note: Award M1 for a weighted average of probabilities with weights 0.1, 0.9 . [2 marks] (d) attempt to use conditional probability formula M1

0.9 0.05

0.108875997

(A1)

0.41334 0.413 A1 [3 marks] Total [10 marks]

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Mathematics Higher level - aptutorgroup.com · Math higher level paper 3 statistics and probability specimen question paper Math higher level paper 3 statistics and probability specimen

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